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Suppose a 87.5-kg sprinter accelerates from rest to 11 m/s in 2.8 s. What is the average acceleration?

a) 1.25m/s²
b) 3.93m/s²
c) 7.86m/s²
d) 20.71m/s²

User Patidati
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1 Answer

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Final answer:

The average acceleration of an 87.5-kg sprinter who goes from rest to 11 m/s in 2.8 seconds is calculated using the formula for acceleration (a = change in velocity over time). The acceleration is determined to be 3.93 m/s², which corresponds to option (b).

Step-by-step explanation:

The student asks about calculating the average acceleration of an 87.5-kg sprinter who accelerates from rest to 11 m/s in 2.8 s. Acceleration is defined as the change in velocity divided by the time over which the change occurs. Therefore, to find the sprinter's acceleration, we use the formula:

a = Δv / t

Where a is the acceleration, Δv is the change in velocity, and t is the time interval.

Since the sprinter accelerates from rest, his initial velocity, v₀, is 0 m/s and his final velocity, v₁, is 11 m/s. The time, t, is given as 2.8 s. Plugging these values into the formula gives:

a = (11 m/s - 0 m/s) / 2.8 s

a = 11 m/s / 2.8 s

a = 3.93 m/s²

The answer is (b) 3.93 m/s², which represents the average acceleration of the sprinter over the 2.8 second period.

User Dajuric
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